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# If j and k are positive integers, j 2 is divisible by 4

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Manager
Joined: 18 Aug 2005
Posts: 85
Location: Italy, Rome caput mundi
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If j and k are positive integers, j 2 is divisible by 4 [#permalink]  27 Sep 2005, 03:15
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If j and k are positive integers, j â€“ 2 is divisible by 4 and k â€“ 5 is divisible by 4, all of the following could be the value of j â€“ k EXCEPT:

43

33

21

13

5
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"To win the battle you need to last 15 minutes longer than your enemies"

VP
Joined: 13 Jun 2004
Posts: 1123
Location: London, UK
Schools: Tuck'08
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Kudos [?]: 31 [0], given: 0

I made 2 lists of potential numbers. I found 43 is not possible.
Senior Manager
Joined: 30 Oct 2004
Posts: 287
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Kudos [?]: 11 [0], given: 0

Yup.. 43 here too.

Here's my explanation

j>0, k>0
Also, j-2 = 4X (some number X since j-2 is a multiple of 4)
k-5 = 4Y (same reasoning)
Combining the two we get

j-k = (X-Y)(4) - 3 => j-k = A - 3 (some number A, which is 4(X-Y))

In the list of answers... add 3 to all... The one that in NOT a multiple of 4 is the answer. 43 + 3 = 46 is not a multiple of 4.
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Manager
Joined: 06 Aug 2005
Posts: 197
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Kudos [?]: 5 [0], given: 0

It is just modulo 4 arithmetic.

(j-2)= 0 mod 4
So j = 2 mod 4

(k-5) = 0 mod 4
So k = 5 mod 4 = 1 mod 4

j-k = (2-1) mod 4 = 1 mod 4

43 = 3 mod 4.
So j-k <> 43
Senior Manager
Joined: 14 Apr 2005
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Location: India, Chennai
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Kudos [?]: 6 [0], given: 0

Re: PS [#permalink]  28 Sep 2005, 04:17
Macedon wrote:
If j and k are positive integers, j â€“ 2 is divisible by 4 and k â€“ 5 is divisible by 4, all of the following could be the value of j â€“ k EXCEPT:

43

33

21

13

5

My answer is A, and my work is as below.
j-2= 0 mod 4; k-5= 0 mod 4
=> j-k= 3 + 0 mod 4 or j-k = 1 mod 4
5 when divided by 4 gives remainder 1,
21 when divided by 4 gives remainder 1,
33 when divided by 4 gives remainder 1,
43 when divided by 4 gives remainder 3, and hence the answer is 43.
Re: PS   [#permalink] 28 Sep 2005, 04:17
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